GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
√(2x + 2) −√(x − 3) = 2
[#permalink]
07 Feb 2022, 07:17
07 Feb 2022, 07:17
3
5
Bookmarks
00:00
Question Stats:
64% (01:32) correct
35% (01:42) wrong based on 28 sessions
Hide Show timer Statistics
\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(x\) |
\(7\) |
A) The quantity in Column A is greater.
B) The quantity in Column B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
√(2x + 2) −√(x − 3) = 2
[#permalink]
07 Feb 2022, 17:06
07 Feb 2022, 17:06
GreenlightTestPrep wrote:
\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
Quantity A |
Quantity B |
\(x\) |
\(7\) |
The equation implies that 2x+2 and x-3 are both PERFECT SQUARES, since the difference between their square roots is an INTEGER.
Further, 2x+2 and x-3 must be two perfect squares that are relatively close to each other, since the difference between their square roots is quite small (2).
Only the value given for Quantity B -- 7 -- will render for 2x+2 and x-3 two perfect squares that are relatively close to each other:
2x+2 = 2*7 + 2 = 16
x-3 = 7-3 = 4
Implication:
x=7 is the only possible solution for the equation, with the result that Quantities A and B are equal.
Show: ::
C
√(2x + 2) −√(x − 3) = 2
[#permalink]
07 Feb 2022, 17:24
07 Feb 2022, 17:24
A useful identity:
\((a+b)^2 = a^2 + 2ab + b^2\)
Algebraic solution:
Adding \(\sqrt{x − 3}\) to both sides, we get:
\(\sqrt{2x + 2}=\sqrt{x − 3} + 2\)
Squaring both sides, we get:
\(2x+2 = (x-3) + (2*\sqrt{x−3}*2) + 2^2\)
\(2x+2 = x+1+4\sqrt{x − 3}\)
Subtracting x+1 from both sides so that the root is more isolated, we get:
\(x+1=4\sqrt{x − 3}\)
Squaring both sides, we get:
\(x^2 + 2*x*1 + 1^2 = 16(x-3)\)
\(x^2 + 2x + 1 = 16x-48\)
\(x^2-14x+49 = 0\)
\((x-7)^2 = 0\)
\(x=7\)
Thus, Quantities A and B are equal.
\((a+b)^2 = a^2 + 2ab + b^2\)
GreenlightTestPrep wrote:
\(\sqrt{2x + 2} −\sqrt{x − 3} = 2\)
Quantity A |
Quantity B |
\(x\) |
\(7\) |
Algebraic solution:
Adding \(\sqrt{x − 3}\) to both sides, we get:
\(\sqrt{2x + 2}=\sqrt{x − 3} + 2\)
Squaring both sides, we get:
\(2x+2 = (x-3) + (2*\sqrt{x−3}*2) + 2^2\)
\(2x+2 = x+1+4\sqrt{x − 3}\)
Subtracting x+1 from both sides so that the root is more isolated, we get:
\(x+1=4\sqrt{x − 3}\)
Squaring both sides, we get:
\(x^2 + 2*x*1 + 1^2 = 16(x-3)\)
\(x^2 + 2x + 1 = 16x-48\)
\(x^2-14x+49 = 0\)
\((x-7)^2 = 0\)
\(x=7\)
Thus, Quantities A and B are equal.
Show: ::
C
